Why does adding up all the entries of an adjoint matrix of a 2n x 2n sized skew-symmetric matrix equal 0

linear algebraskew-symmetric matrices

I'm in the middle of proving this statement.

If $A$ is a $2n \times 2n$ sized skew-symmetric matrix, then $det(A) = det(A + xJ)$ whereas $J$ is an $2n \times 2n$ sized matrix with all of its entries 1.

Since it is self-evident that the statement is true when $x = 0$, I tried to prove the statement when $x \neq 0$.

Let $\boldsymbol{j}$ be a $1 \times 2n$ row vector with all its entries 1.

$|A + xJ| = \begin{vmatrix}A_1+x\boldsymbol{j} \\ A_2 + x\boldsymbol{j}\\\vdots\\A_{2n}+x\boldsymbol{j}\end{vmatrix} = \begin{vmatrix}A_1 \\ A_2 + x\boldsymbol{j}\\\vdots\\A_{2n}+x\boldsymbol{j}\end{vmatrix} + \begin{vmatrix}x\boldsymbol{j} \\ A_2 + x\boldsymbol{j}\\\vdots\\A_{2n}+x\boldsymbol{j}\end{vmatrix} = \begin{vmatrix}A_1 \\ A_2 + x\boldsymbol{j}\\\vdots\\A_{2n}+x\boldsymbol{j}\end{vmatrix} + x\begin{vmatrix}\boldsymbol{j} \\ A_2\\\vdots\\A_{2n}\end{vmatrix} = \cdots = |A| + x\Sigma^{2n}_{i=1} |A_i(\boldsymbol{j})|$

Since $|A_i(\boldsymbol{j})|$ equals the sum of all entries of adjoint matrix of A, the statement is now equivalent to the title. So.. how do I prove that adding up all the entries of an adjoint matrix of a 2n x 2n sized skew-symmetric matrix equal 0?

Best Answer

In this answer, we require, as part of the definition, that a skew-symmetric matrix must be hollow, i.e. it must possess a zero diagonal. If the characteristic of the underlying field is not $2$, this requirement is redundant because it follows from the condition that $A^T=-A$. However, when the field has characteristic $2$, this hollowness requirement is genuinely additional.

In general, when $K$ is a skew-symmetric matrix, the rank of $K$ must be even and $x^TKx=0$ for any vector $x$. In your case, suppose that $A$ is an even-sized skew-symmetric matrix. Denote by $M_{ij}$ the submatrix obtained by deleting the $i$-th row and the $j$-th column of $A$. We have two observations:

  1. Each principal submatrix $M_{ii}$ is skew-symmetric. Therefore $\operatorname{rank}(M_{ii})$ is even. However, as the size of $M_{ii}$ is odd, $M_{ii}$ must be singular. Hence $\det M_{ii}=0$, i.e. $\operatorname{adj}(A)$ is hollow.
  2. When $i\ne j$, we have $M_{ij}=-M_{ji}^T$. As the sizes of $M_{ij}$ and $M_{ji}$ are odd, $\det M_{ij}=-\det M_{ji}$. Hence the off-diagonal part of $\operatorname{adj}(A)$ is skew-symmetric.

It follows that $\operatorname{adj}(A)$ is skew-symmetric. (If the size of $A$ is odd instead, then $\operatorname{adj}(A)$ will be symmetric rather than skew-symmetric). Therefore, when $e=(1,1,\ldots,1)^T$, we have $e^T\operatorname{adj}(A)e=0$, i.e. the sum of all entries of $\operatorname{adj}(A)$ is equal to zero. It follows that $\det(A+ee^T)=\det(A)+e^T\operatorname{adj}(A)e=\det(A)$.

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